Question

Find area of region common to circle x^2+y^2=16 and the parabola y^2=6x about y axis

Answer 1

Answer:

Toolbox:

Suppose we are given two curves represented by y=f(x),y=g(x) where f(x)≥g(x) in [a.b] the points of intersection of the two curves are given by x=a and x=b by taking common values of y from the given equation of the curves.

Hence the required area is given by A=∫ba[f(x)−g(x)]dx

Step 1:

The required area is area of the circle x2+y2=16 which is extension to the parabola y2=6x

To find the points of intersection,let us solve the given equation.

substituing for y2 in the equation of the circle we get,

x2+6x=16

⇒x2+6x−16=0

on factorising we get ,

(x+8)(x−2)=0

⇒x=−8andx=2.

Step 2:

since for x=-8 we get y2=−8 which is imaginary,we cannot take this value.

Now if x=2 we get y=±23–√.

Hence the points of intersection are (2,23–√) and (2,−23–√).

Now let us take the limits as 0 to 2 for parabola and 2 to 4 for the circle.

The area bounded by the circle and parabola is 2x[area bounded by the parabola and x-axis +area bounded by the circle and x-axis] A=2×[∫206x−−√dx+∫4216−x2−−−−−−√dx]

on integrating we get

A=2[6–√x3232]20+x2[16−x2−−−−−−√+162sin−1x4]42

Step 3:

on applying limits we get,

A=46–√3(22–√)+2[0+162sin1(1)−16−4−−−−−√+162sin−112]

A=46–√3(22–√)+2[8π2−43–√+8(π6)]

on simplifying we get,

A=43[43–√+6π−33–√−2π]

=43[3–√+4π]

Area of the circle=πr2

⇒π(4)2

⇒16π sq.units.

The required area =16π−43[4π+3–√]

⇒43[4×3π−4π−3–√]

⇒43[8π−3–√]sq.units.

Hence the required area=43[8π−3–√]sq.units.

how become limits 0 to 2 for parabola and 2 to 4 for the circle..can u help me